Density of the gas is 3.05 × 10⁻³ g / cm³.
<u>Explanation:</u>
Volume of the cylinder = π r² h
where r is the radius and h is the height of the height or the length of the glass tube.
Here r = 4 cm and h = 27.4 cm
Volume of the cylinder = 3.14 × 4 × 4 × 27.4 = 1376.6 cm³
We have to find the mass of the gas by subtracting the mass of the tube filled with the substance from the mass of the empty tube.
Mass of the substance = 258.5 - 254.3 = 4.2 g
We have to find the density using the formula as,
Plugin the values as,
= 3.05 × 10⁻³ g / cm³
So the Density of the gas is 3.05 × 10⁻³ g / cm³.
Answer:
Explanation:
Hello there!
In this case, according to the given information it will be firstly necessary to set up the chemical equation taking place:
We infer we need to calculate the moles of NH3 by using both of the moles of N2 and H2 at the beginning, in order to identify the limiting reactant:
Thus, since hydrogen yields the fewest moles of ammonia, we conclude that we are just able to yield 4 moles of NH3.
Regards!
Cs+1
The only common oxidation state is +1.
setup 1 : to the right
setup 2 : equilibrium
setup 3 : to the left
<h3>Further explanation</h3>
The reaction quotient (Q) : determine a reaction has reached equilibrium
For reaction :
aA+bB⇔cC+dD
![\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Comparing Q with K( the equilibrium constant) :
K is the product of ions in an equilibrium saturated state
Q is the product of the ion ions from the reacting substance
Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)
Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium
Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)
Keq = 6.16 x 10⁻³
Q for reaction N₂O₄(0) ⇒ 2NO₂(g)
![\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
Setup 1 :
Q<K⇒The reaction moved to the right (products)
Setup 2 :
Q=K⇒the system at equilibrium
Setup 3 :
Q>K⇒The reaction moved to the left (reactants)
Physical change is the color, shape stuff like that and chemical change is once it’s changed it can’t go back
